BÀi 3:
1: \(\left(x^2-4x\right)^2+2\left(x-2\right)^2=43\)
=>\(\left(x^2-4x+4-4\right)^2+2\left(x-2\right)^2=43\)
=>\(\left\lbrack\left(x-2\right)^2-4\right\rbrack^2+2\left(x-2\right)^2-43=0\)
=>\(\left(x-2\right)^4-8\left(x-2\right)^2+16+2\left(x-2\right)^2-43=0\)
=>\(\left(x-2\right)^4-6\left(x-2\right)^2-27=0\)
=>\(\left\lbrack\left(x-2\right)^2-9\right\rbrack\left\lbrack\left(x-2\right)^2+3\right\rbrack=0\)
=>\(\left(x-2\right)^2-9=0\)
=>\(\left(x-2\right)^2=9\)
=>\(\left[\begin{array}{l}x-2=3\\ x-2=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=-1\end{array}\right.\)
2:
ĐKXĐ: x∉{-4;-6;-2;-8}
\(\frac{x^2+8x+20}{x+4}+\frac{x^2+12x+42}{x+6}=\frac{x^2+4x+6}{x+2}+\frac{x^2+16x+72}{x+8}\)
=>\(\frac{x^2+8x+16+4}{x+4}+\frac{x^2+12x+36+6}{x+6}=\frac{x^2+4x+4+2}{x+2}+\frac{x^2+16x+64+8}{x+8}\)
=>\(\frac{\left(x+4\right)^2+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}=\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}\)
=>\(x+4+\frac{4}{x+4}+x+6+\frac{6}{x+6}=x+2+\frac{2}{x+2}+x+8+\frac{8}{x+8}\)
=>\(\frac{4}{x+4}+\frac{6}{x+6}=\frac{2}{x+2}+\frac{8}{x+8}\)
=>\(\frac{4-x-4}{x+4}+\frac{6-x-6}{x+6}=\frac{2-x-2}{x+2}+\frac{8-x-8}{x+8}\)
=>\(-x\cdot\left(\frac{1}{x+4}+\frac{1}{x+6}-\frac{1}{x+2}-\frac{1}{x+8}\right)=0\)
TH1: -x=0
=>x=0(nhận)
TH2: \(\frac{1}{x+4}+\frac{1}{x+6}-\frac{1}{x+2}-\frac{1}{x+8}=0\)
=>\(\frac{x+4+x+6}{\left(x+4\right)\left(x+6\right)}-\frac{x+2+x+8}{\left(x+2\right)\left(x+8\right)}=0\)
=>\(\left(2x+10\right)\left(\frac{1}{\left(x+4\right)\left(x+6\right)}-\frac{1}{\left(x+2\right)\left(x+8\right)}\right)=0\)
=>\(\left[\begin{array}{l}2x+10=0\\ \left(x+4\right)\left(x+6\right)=\left(x+2\right)\left(x+8\right)\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=-10\\ x^2+10x+24=x^2+10x+16\end{array}\right.\)
=>2x=-10
=>x=-5(nhận)
Bài 2:
1: \(x^2-2xy+y^2-4x+4y-5\)
\(=\left(x-y\right)^2-4\left(x-y\right)-5\)
\(=\left(x-y\right)^2-5\left(x-y\right)+\left(x-y\right)-5\)
=(x-y-5)(x-y+1)
